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# Official Quant Thread for CAT 2012 [Part 1]

Hi puys, CAT 2011 season is finally over :: :: Let start a new season of Quants thread for CAT 2012.. :: :: Here are few links of Quants thread for CAT 2011 season... Part-8 http://www.pagalguy.com/forum/quanti...at-...
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A square tin sheet of side 12 inches is converted into a box with open top in the following steps : The sheet is placed horizontally. Then, equal sized squares, each of side x inches, are cut from the four corners of the sheet. Finally, the four resulting sides are bent vertically upwards in the shape of a box. If x is an integer, then what value of x maximizes the volume of the box

a.) 3...........b.) 4...........c.) 1...........d.) 2...........e.) 5

Answer is d.) 2
Education is what remains after one has forgotten what one has learned in school
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bhai let centre of circle be x,y
x^2+y^2=(x-1)^2+y^2
x=1/2
similarly y=1/2
ab k=0 or k=5/13 aaenge but k=0 will make points same so rejected
therefore one value of k=5/13
bhai ko fever main bhi chenn nhi h....
thats why we hail varun bhai....

bhai it is x=2........
(12-x)^2.x
differentiate and make maximum

:cheerio: both are correct......
Bhai abhi dawai leke thoda sa theek feel kar raha hoon plus din bhar sone se ab neend bhi ni aa rahi to socha apne Puylog se he mil leta hoon :D

Q.) Both roots of the quadratic equation x^2 - 63x + k = 0 are prime numbers. The number of possible values of k is
a.) 0..........b.) 1........c.) 2.........d.) 4.........e.) more than 4
Smoothest seas do not make tough sailors
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Q) Four distinct points (2k, 3k), (1, 0), (0, 1) and (0, 0) lie on the same circle for:

a.) for one value of k such that k 1
b.) for two values of k such that k c.) for one value of k such that 0 k 1
d.) for one value of k such that 0 k e.) Cannot be determined

x2 + y2 + 2gx + 2fy + c = 0
1 + 2g + c = 0

similarly
1 + 2f + c = 0

similarly c = 0

center = 1/2,1/2

radius = sqrt(1/4+1/4) = sqrt(1/2)

4k2 + 9k2 - 2k - 3k = 0

13k2 -5k = 0

k = 0 or 5/13

hence 1 value...

for k = 0 .. point circle will exist
• 2 Likes
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bhai plz check aapke method se bhi numbers will be
21
21x2
21x3
21x4
lcm will be 21x3x4=252 jo bahot zada h.....

yeah,u're right buddy..Thanks for pointing out
I won't run on Sunday even if it's 100m event in Olympic games
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A square tin sheet of side 12 inches is converted into a box with open top in the following steps : The sheet is placed horizontally. Then, equal sized squares, each of side x inches, are cut from the four corners of the sheet. Finally, the four resulting sides are bent vertically upwards in the shape of a box. If x is an integer, then what value of x maximizes the volume of the box

a.) 3...........b.) 4...........c.) 1...........d.) 2...........e.) 5

bhai it is x=2........
(12-x)^2.x
differentiate and make maximum
• 1 Like
so many gals hv lovd me,so many hv left me,still no1 was btr enuf dn u my quant ! ur love
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Q) Four distinct points (2k, 3k), (1, 0), (0, 1) and (0, 0) lie on the same circle for:

a.) for one value of k such that k 1
b.) for two values of k such that k c.) for one value of k such that 0 k 1
d.) for one value of k such that 0 k e.) Cannot be determined

bhai let centre of circle be x,y
x^2+y^2=(x-1)^2+y^2
x=1/2
similarly y=1/2
ab k=0 or k=5/13 aaenge but k=0 will make points same so rejected
therefore one value of k=5/13
bhai ko fever main bhi chenn nhi h....
thats why we hail varun bhai....
• 2 Likes
so many gals hv lovd me,so many hv left me,still no1 was btr enuf dn u my quant ! ur love
Commenting on this post has been disabled.

A square tin sheet of side 12 inches is converted into a box with open top in the following steps : The sheet is placed horizontally. Then, equal sized squares, each of side x inches, are cut from the four corners of the sheet. Finally, the four resulting sides are bent vertically upwards in the shape of a box. If x is an integer, then what value of x maximizes the volume of the box

a.) 3...........b.) 4...........c.) 1...........d.) 2...........e.) 5

Smoothest seas do not make tough sailors
Commenting on this post has been disabled.

Q) Four distinct points (2k, 3k), (1, 0), (0, 1) and (0, 0) lie on the same circle for:

a.) for one value of k such that k 1
b.) for two values of k such that k c.) for one value of k such that 0 k 1
d.) for one value of k such that 0 k e.) Cannot be determined

Smoothest seas do not make tough sailors
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bhai options were 42 84 60 105...

mast approach kya hain

I don't know I am getting a bit messed up
still I am posting this..

so numbers are a, b, c, d and a, b, c should be factor of d

as 210= 2*3*5*7

so it has to be multiple of 7

so I can say 7x+7x+7x+ 2*7x= 35x =>210

x=6

so total = 42, 42, 42, 84

so 84 is the answer
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bhai options were 42 84 60 105...

mast approach kya hain

bhai mast approach jaldi main question krne k lie h
in a,b,c,d ko equal kar do to a=105/2 ata h yani 52.5
ab lcm isse kam nhi hoga bt iske aaspass hi hoga
to bas 1 hi option h iske pass 60...
• 1 Like
so many gals hv lovd me,so many hv left me,still no1 was btr enuf dn u my quant ! ur love
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abhishek 2011 Says
allan bhai aap 1 bar is question k option post kae sakte ho...plz....option se karna easy rahega shayad 1 achi approach h mere pass....

bhai options were 42 84 60 105...

mast approach kya hain
IIM Kozhikode PGP 2014-16| CAT'13 99.43 | XAT'14 97.21  http://www.pagalguy.com/discussions/all-i-wanted-to-speak-about-cat-25002933/19890122
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Assuming numbers to be distinct,we have to maximise the HCF,let's say a = Hk,b=Hl,c=Hm,d= Hn where H is HCF and k,l,m,n different natural#'s

H(k+l+m+n) = 210,

now k,l,m,n have to be different with least possible values and H has to be maximum

H = 21 and k+l+m+n = 10 would be feasible here...

bhai plz check aapke method se bhi numbers will be
21
21x2
21x3
21x4
lcm will be 21x3x4=252 jo bahot zada h.....
• 2 Likes
so many gals hv lovd me,so many hv left me,still no1 was btr enuf dn u my quant ! ur love
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lcm will be minimum if all numbers are equal ..

but it is not div by 4

so we take 3 equal and 1 not equal

3ax + bx = 210
x(3a+b) = 210

x = 30

3a + b = 7
a = 2
b = 1

numbers -- 60,60,60,30

shree bhai bilkul sahi baat
if the number had been 212 lcm wud have been 53
bt in this case putting a,b,c,d equal gives 105/2
so lcm has to be lesser than 62.5 thats why i was asking for options in the above post
isse jst cum vala answer hoga.....options main...method can be used to eliminate options in case kam time main fluke marna ho to....
• 1 Like
so many gals hv lovd me,so many hv left me,still no1 was btr enuf dn u my quant ! ur love
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Shawn90 Says
bhai hit and trial chor ke koi dusra approach ho sakta hain???

allan89 Says
oa is 60....is there any other way to do these than hit and trial??

Actually this is calculation only more than hit and tiral. But since its trivial, I said it was h&t.;
Lets see.

As this is clear that Lcm would be minimum when 3 of the 4 numbers are a factor of the third.
Let a b and c be factors of d.
This makes d the largest.

Case 1. -> All of a b and c are 1/2 of d (as they cannot be more than that).
=> 1/2d + 1/2d + 1/2d + d ==> 5/2*d >= 210
=> d >= 210.

Case 2. -> When any one of a b or c is equal to d.
Let a = d.
=> 1/2d + 1/2d + d + d ==> 3d >= 210.
=> d >= 70.

Case 3. -> When any two of a b or c are equal to d.
Let a = b = d.
=> 1/2d + d + d +d ==> 7/2*d >= 210.
=> d >= 60.

Clearly Case 3 gives us the minimum possible value of d (and hence the Lcm).
Since we took a = b = d.
=> a = b = d = 60.
and c = 30.

Hope it is clear now.
• 2 Likes
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allan89 Says
what can be the minimum lcm of 4 ns a,b,c,d whose sum is 210?

lcm will be minimum if all numbers are equal ..

but it is not div by 4

so we take 3 equal and 1 not equal

3ax + bx = 210
x(3a+b) = 210

x = 30

3a + b = 7
a = 2
b = 1

numbers -- 60,60,60,30
• 3 Likes
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